Carbonyls & Carboxylic Acids

Which observation distinguishes propanal from propanone when each is warmed with Fehling's solution?

• ABoth give a silver mirror
• BBoth decolourise the blue solution to colourless
• CPropanal gives a brick-red precipitate; propanone shows no change✓
• DPropanone gives a brick-red precipitate; propanal shows no change

Answer: Fehling's solution oxidises aldehydes only. Propanal (an aldehyde) reduces the blue Cu²⁺ complex to a brick-red precipitate of copper(I) oxide, Cu₂O; propanone (a ketone) cannot be oxidised so the solution stays blue.

An aldehyde RCHO is oxidised to a carboxylic acid using acidified potassium dichromate(VI). What is the colour change observed?

• AColourless to pink
• BOrange to green✓
• CPurple to colourless
• DGreen to orange

Answer: Dichromate(VI), Cr₂O₇²⁻, is orange and is reduced to Cr³⁺ which is green. The aldehyde is oxidised to a carboxylic acid: RCHO + [O] → RCOOH.

Butanone is reduced using NaBH₄ in aqueous ethanol. What is the organic product, and how is NaBH₄ best described as a reducing agent here?

• AButan-1-ol; NaBH₄ delivers a hydride ion (H⁻) nucleophile
• BButan-2-ol; NaBH₄ delivers a hydride ion (H⁻) nucleophile✓
• CButan-2-ol; NaBH₄ delivers a proton (H⁺) electrophile
• DButanal; NaBH₄ delivers a hydrogen radical

Answer: NaBH₄ provides a hydride ion (H⁻) which acts as a nucleophile, attacking the carbonyl carbon. Reduction of the ketone butanone (CH₃COCH₂CH₃) gives the secondary alcohol butan-2-ol.

Ethanal reacts with HCN in the presence of a trace of KCN to form 2-hydroxypropanenitrile. What is the role of the cyanide ion, and why is a trace of KCN needed?

• ACN⁻ is an electrophile; KCN provides H⁺ to start the reaction
• BCN⁻ removes the carbonyl oxygen; KCN acts as an oxidising agent
• CCN⁻ is a catalyst that is not consumed; KCN provides the carbon skeleton
• DCN⁻ is the nucleophile that attacks the carbonyl carbon; KCN raises the concentration of the CN⁻ nucleophile✓

Answer: The mechanism is nucleophilic addition. The cyanide ion CN⁻ attacks the δ+ carbonyl carbon. HCN is a weak acid (low [CN⁻]), so a trace of KCN increases the concentration of the CN⁻ nucleophile and speeds the reaction.

Tollens' reagent is the ammoniacal silver nitrate solution. Which species is the active oxidising agent, and what is the positive result with an aldehyde?

• AAg⁺ (as [Ag(NH₃)₂]⁺); a silver mirror forms✓
• BNH₃; a white precipitate forms
• CNO₃⁻; a brown gas is evolved
• DCu²⁺; a brick-red precipitate forms

Answer: Tollens' reagent contains the complex [Ag(NH₃)₂]⁺. An aldehyde reduces the silver(I) to metallic silver, depositing a silver mirror on the tube wall, while the aldehyde is oxidised to a carboxylate ion.

Which compound gives a positive (yellow precipitate) result in the triiodomethane (iodoform) test with alkaline aqueous iodine?

• APropanal, CH₃CH₂CHO
• BPentan-3-one, CH₃CH₂COCH₂CH₃
• CPropan-2-ol, CH₃CH(OH)CH₃✓
• DMethanol, CH₃OH

Answer: The iodoform test is positive for compounds containing a CH₃CO– group OR a CH₃CH(OH)– group (which is oxidised in situ to CH₃CO–). Propan-2-ol contains CH₃CH(OH)– and gives a pale-yellow precipitate of CHI₃. Pentan-3-one and propanal lack the required CH₃CO– group.

Why is chloroethanoic acid (ClCH₂COOH, Ka = 1.3 × 10⁻³) a stronger acid than ethanoic acid (CH₃COOH, Ka = 1.7 × 10⁻⁵)?

• AThe chlorine atom donates electron density, strengthening the O–H bond
• BChloroethanoic acid forms more hydrogen bonds with water
• CThe electron-withdrawing chlorine stabilises the carboxylate anion by delocalising its negative charge, favouring dissociation✓
• DChlorine increases the molar mass, which raises acidity

Answer: The electronegative Cl atom withdraws electron density (inductive –I effect), which disperses and stabilises the negative charge on the ClCH₂COO⁻ anion. The more stable the conjugate base, the more the equilibrium favours dissociation, so Ka is larger.

What are the products when ethanoic acid reacts with solid sodium carbonate?

• ASodium ethoxide and carbon dioxide
• BSodium ethanoate, water and carbon dioxide✓
• CEthanol, water and carbon dioxide
• DSodium ethanoate and hydrogen only

Answer: Carboxylic acids are strong enough acids to liberate CO₂ from carbonates: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. Effervescence of CO₂ is a test that distinguishes carboxylic acids from phenols (phenols do not react with carbonates).

Ethyl ethanoate is heated under reflux with aqueous sodium hydroxide. What does this alkaline (saponification) hydrolysis produce?

• AEthanoic acid and ethanol
• BEthanoic acid and sodium ethoxide
• CSodium ethanoate and ethanol✓
• DEthanol and water only

Answer: Alkaline hydrolysis of an ester is irreversible and gives the carboxylate salt plus the alcohol: CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH. (Acid hydrolysis would instead give the carboxylic acid CH₃COOH and the alcohol, and is reversible.)

Which combination of reactants forms the ester propyl methanoate, and what is the standard condition for this esterification?

• AMethanol + propanoic acid, concentrated H₂SO₄ catalyst
• BMethanoic acid + propanal, concentrated H₂SO₄ catalyst
• CEthanoic acid + propan-1-ol, NaOH catalyst
• DMethanoic acid + propan-1-ol, concentrated H₂SO₄ catalyst✓

Answer: An ester R-COO-R' is named acyl-from-acid + alkyl-from-alcohol. Propyl methanoate (HCOOCH₂CH₂CH₃) comes from methanoic acid (HCOOH, the acyl part) and propan-1-ol (the propyl part), warmed with a concentrated sulfuric acid catalyst.

Ethanoyl chloride (CH₃COCl) is added to water. What are the products and the key observation?

• AEthanol and chlorine; a green gas is seen
• BEthanoic acid and HCl; steamy/misty white fumes of HCl are seen✓
• CEthanoic anhydride and water; no visible change
• DEthanal and HCl; a silver mirror forms

Answer: Acyl chlorides react vigorously with water: CH₃COCl + H₂O → CH₃COOH + HCl. The HCl produced is seen as steamy/misty white fumes, a characteristic test for an acyl chloride.

Ethanoic anhydride [(CH₃CO)₂O] is used industrially in preference to ethanoyl chloride to make aspirin and other esters/amides. For its reaction with an amine RNH₂, which product and advantage are correct?

• AIt gives RNH₃⁺Cl⁻ + CH₃CHO; it avoids forming an amide
• BIt gives RN(COCH₃)₂ + H₂O; it is a stronger oxidising agent
• CIt gives RNHCOCH₃ + HCl; it reacts faster than the acyl chloride
• DIt gives RNHCOCH₃ + CH₃COOH; it is cheaper, less corrosive and does not release HCl fumes✓

Answer: An acid anhydride acylates an amine to give an N-substituted amide plus a carboxylic acid: (CH₃CO)₂O + RNH₂ → CH₃CONHR + CH₃COOH. It is preferred over the acyl chloride because it is cheaper, less corrosive/violent, less readily hydrolysed by moisture, and does not produce corrosive HCl fumes.